Question 1 |
The transfer function of a phase lead compensator is given by
D(s)=\frac{3\left ( s+\frac{1}{3T} \right )}{\left ( s+\frac{1}{T} \right )}
The frequency (in rad/sec), at which \angle D(j\omega ) is maximum, is
D(s)=\frac{3\left ( s+\frac{1}{3T} \right )}{\left ( s+\frac{1}{T} \right )}
The frequency (in rad/sec), at which \angle D(j\omega ) is maximum, is
\sqrt{\frac{3}{T^2}} | |
\sqrt{\frac{1}{3T^2}} | |
\sqrt{3T} | |
\sqrt{3T^3} |
Question 1 Explanation:
T(s)=\frac{1+3Ts}{1+Ts}
Frequency at which \angle T(j\omega ) is maximum,
(\omega _m)=\frac{1}{T\sqrt{\alpha }}
\alpha =\frac{1}{1/3}=3
\omega _m=\frac{1}{T\sqrt{3}}=\sqrt{\frac{1}{3T^2}}
Frequency at which \angle T(j\omega ) is maximum,
(\omega _m)=\frac{1}{T\sqrt{\alpha }}
\alpha =\frac{1}{1/3}=3
\omega _m=\frac{1}{T\sqrt{3}}=\sqrt{\frac{1}{3T^2}}
Question 2 |
The transfer function C(s) of a compensator is given below.
C(s)=\frac{(1+\frac{s}{0.1})(1+\frac{s}{100})}{(1+s)(1+\frac{s}{10})}
The frequency range in which the phase (lead) introduced by the compensator reaches the maximum is
C(s)=\frac{(1+\frac{s}{0.1})(1+\frac{s}{100})}{(1+s)(1+\frac{s}{10})}
The frequency range in which the phase (lead) introduced by the compensator reaches the maximum is
0.1 \lt \omega \lt 1 | |
1 \lt \omega \lt 10 | |
10 \lt \omega \lt 100 | |
\omega \gt 100 |
Question 2 Explanation:
Pole zero diagram of compensator transfer function is shown below.
Maximum phase lead is between 0.1 and 1.
0.1 \lt \omega \lt 1
Maximum phase lead is between 0.1 and 1.
0.1 \lt \omega \lt 1
Question 3 |
For the network shown in the figure below, the frequency (in rad/s) at which the maximum phase lag occurs is, ___________.
0.15 | |
0.32 | |
0.66 | |
0.92 |
Question 3 Explanation:
Assuming
\begin{aligned} R_1T&=9\Omega \\ R_2&=1\Omega\\ \text{ We can write,} \\ \frac{V_o(s)}{V_{in}(s)}&=\frac{R_2+\frac{1}{sC}}{R_1+R_2+\frac{1}{sC}} \\ &=\frac{1+R_2\cdot Cs}{1+(R_1+R_2)Cs} \\ &=\frac{1+R_2\cdot Cs}{1+\left (\frac{R_1+R_2}{R_2} \right )R_2 Cs}\\ \text{Let, } R_2C&=T \\ \frac{R_1+R_2}{R_2}&=\beta \\ \text{Hence, } \\ \frac{V_o(s)}{V_{in}(s)}&=\frac{1+Ts}{(1+\beta Ts)} \\ \text{Which represent }& \text{a lag compensator}\\ \because \text{ Here, } T&=R_2C=1.1=1 sec \\ \beta &=\frac{1+9}{1}=10\\ \text{Maximum phase }& \text{lag occurs at frequency}\\ \omega _n&=\frac{1}{T\sqrt{\beta }} \\ &=\frac{1}{1\sqrt{10}} =0.316 rad/sec. \end{aligned}
\begin{aligned} R_1T&=9\Omega \\ R_2&=1\Omega\\ \text{ We can write,} \\ \frac{V_o(s)}{V_{in}(s)}&=\frac{R_2+\frac{1}{sC}}{R_1+R_2+\frac{1}{sC}} \\ &=\frac{1+R_2\cdot Cs}{1+(R_1+R_2)Cs} \\ &=\frac{1+R_2\cdot Cs}{1+\left (\frac{R_1+R_2}{R_2} \right )R_2 Cs}\\ \text{Let, } R_2C&=T \\ \frac{R_1+R_2}{R_2}&=\beta \\ \text{Hence, } \\ \frac{V_o(s)}{V_{in}(s)}&=\frac{1+Ts}{(1+\beta Ts)} \\ \text{Which represent }& \text{a lag compensator}\\ \because \text{ Here, } T&=R_2C=1.1=1 sec \\ \beta &=\frac{1+9}{1}=10\\ \text{Maximum phase }& \text{lag occurs at frequency}\\ \omega _n&=\frac{1}{T\sqrt{\beta }} \\ &=\frac{1}{1\sqrt{10}} =0.316 rad/sec. \end{aligned}
Question 4 |
The transfer function of a compensator is given as
G_c(s)=\frac{s+a}{s+b}
The phase of the above lead compensator is maximum at
G_c(s)=\frac{s+a}{s+b}
The phase of the above lead compensator is maximum at
\sqrt{2} rad/s | |
\sqrt{3} rad/s | |
\sqrt{6} rad/s | |
1/ \sqrt{3} rad/s |
Question 4 Explanation:
\omega _m=\sqrt{\omega _{c1} \times \omega _{c2}}
\;\;=\sqrt{1 times 2}=\sqrt{2} rad/sec.
\;\;=\sqrt{1 times 2}=\sqrt{2} rad/sec.
Question 5 |
The transfer function of a compensator is given as
G_c(s)=\frac{s+a}{s+b}
G_c(s) is a lead compensator if
G_c(s)=\frac{s+a}{s+b}
G_c(s) is a lead compensator if
a = 1, b = 2 | |
a = 3, b = 2 | |
a =- 3, b =- 1 | |
a = 3, b = 1 |
Question 5 Explanation:
G_c(s)=\frac{s+a}{s+b}
G_c(j\omega )=\frac{j\omega +a}{j\omega +b}
\angle G_c(j\omega )=tan^{-1}\frac{\omega }{a}-tan^{-1}\frac{\omega }{b}
\;\;\;= tan^{-1}\left ( \frac{\frac{\omega }{a}-\frac{\omega }{b}}{1+\frac{\omega ^2 }{ab}} \right )
For G_c(s) to be a lead compensator
\angle G_c(j\omega ) \gt 0
\frac{\omega }{a} \gt \frac{\omega }{b}
\Rightarrow b \gt a
Option (A) Satisfies the above equation.
G_c(j\omega )=\frac{j\omega +a}{j\omega +b}
\angle G_c(j\omega )=tan^{-1}\frac{\omega }{a}-tan^{-1}\frac{\omega }{b}
\;\;\;= tan^{-1}\left ( \frac{\frac{\omega }{a}-\frac{\omega }{b}}{1+\frac{\omega ^2 }{ab}} \right )
For G_c(s) to be a lead compensator
\angle G_c(j\omega ) \gt 0
\frac{\omega }{a} \gt \frac{\omega }{b}
\Rightarrow b \gt a
Option (A) Satisfies the above equation.
Question 6 |
The transfer functions of two compensators are given below :
C_{1}=\frac{10(s+1)}{(s+10)},C_{2}=\frac{s+10}{10(s+1)}
Which one of the following statements is correct ?
C_{1}=\frac{10(s+1)}{(s+10)},C_{2}=\frac{s+10}{10(s+1)}
Which one of the following statements is correct ?
C_{1} is a lead compensator and C_{2} is a lag compensator | |
C_{1} is a lag compensator and C_{2} is a lead compensator | |
Both C_{1}\; and \; C_2 are lead compensator | |
Both C_{1}\; and \; C_2 are lag compensator |
Question 6 Explanation:
C_1=\frac{10(s+1)}{(s+10)}
Zero at s=-1
Pole at s=-10
As zero is closer origin, zero dominates pole. Hence C_1 is lead compensator.
C_2=\frac{s+10}{10(s+1)}
Zero at s=-10
Pole at s=-1
As pole is closer to origin, pole dominates zero. Hence C_2 is lag compensator.
Zero at s=-1
Pole at s=-10
As zero is closer origin, zero dominates pole. Hence C_1 is lead compensator.
C_2=\frac{s+10}{10(s+1)}
Zero at s=-10
Pole at s=-1
As pole is closer to origin, pole dominates zero. Hence C_2 is lag compensator.
Question 7 |
The system \frac{900}{s(s+1)(s+9)} is to be compensated such that its gain-crossover frequency
becomes same as its uncompensated phase crossover frequency and provides a 45^{\circ} phase margin. To achieve this, one may use
a lag compensator that provides an attenuation of 20 dB and a phase lag of 45^{\circ} at the frequency of 3 \sqrt{3} rad/s | |
a lead compensator that provides an amplification of 20 dB and a phase lead of 45^{\circ} at the frequency of 3 rad/s | |
a lag-lead compensator that provides an attenuation of 20 dB and phase
lag of 45^{\circ} at the frequency of \sqrt{3} rad/s | |
a lag-lead compensator that provides an attenuation of 20 dB and phase
lead of 45^{\circ} at the frequency of 3 rad/s |
Question 7 Explanation:
Let uncompensated syatem,
T(s)=\frac{900}{s(s+1)(s+9)}
Phase crossover frequency of uncompensated system =(\omega _{pc}), at this frequency phase of T(j \omega ) is -180^{\circ}
Put s=j\omega \text{ in } T(s)
\begin{aligned} T(j\omega )&=\frac{900}{j\omega (j\omega +1)(j\omega +9)} \\ \angle T(j\omega )&=-90^{\circ}-tan ^{-1}\omega -tan^{-1}\left ( \frac{\omega }{9} \right ) \\ \text{At, } & (\omega _{pc})_1,\angle T(j\omega )=-180^{\circ}\\ -180^{\circ}&=-90^{\circ}-tan^{-1}(\omega _{pc})_1 -tan^{-1}\left ( \frac{(\omega _{pc})_1}{9} \right )\\ -90^{\circ}&=tan^{-1}\frac{(\omega _{pc})_1+\frac{(\omega _{pc})_1}{9}}{1-\frac{(\omega _{pc})^2_1}{9}} \\ \Rightarrow 1-\frac{(\omega _{pc})^2_1}{9}&=0 \\ \Rightarrow (\omega _{pc})_1&= 3 \text{ rad/sec} \end{aligned}
Gain cross frequency of compensated system, (\omega _{gc})_2= phase cross frequency of uncompensated system, (\omega _{pc})_1
\Rightarrow \;\;(\omega _{gc})_2=(\omega _{gc})_1=\; 3 rad/sec
Phase-margin =180^{\circ}+\angle T(j\omega )|_{\omega =(\omega _{gc})_2}
\Rightarrow \; \; \; 45^{\circ}=180^{\circ}+\angle T(j\omega )|_{\omega =(\omega _{pc})_2}
At (\omega _{gc})_2=3 rad/sec,
phase angle of \angle T(j\omega ) is -135^{\circ} and phase of uncompensated systen is -180^{\circ} at 3 rad/sec.
Therefore, the compensator provides phase lead of 45^{\circ} at the frequency of 3 rad/sec.
Let X dB is the gain provided by the compensator, so at gain cross frequency, |T(j\omega )|_{com}=1 or 0 dB. Gain of uncompensated system =|T(j\omega )|_{un-com} =\frac{100}{\omega \sqrt{1+\omega ^2}\sqrt{1+\left ( \frac{\omega }{9} \right )^2}}
|T(j\omega )|_{un-com} in dB
\;=40-20 \log \omega -20 \log \sqrt{1+\omega ^2} -20 \log \sqrt{1+\left ( \frac{\omega }{9} \right )^2}
Gain of compensated system,
|T(j\omega )|_{com}=X+|T(j\omega )_{un-com}|
|T(j\omega )|_{com} must be zero at gain cross frequency (\omega _{gc})_2
|T(j\omega_{gc} )_2|_{com}=X+40-20 \log (\omega _{gc})_2 -20 \log \sqrt{1+(\omega _{gc})^2_2}-20 \log \sqrt{1+\frac{(\omega _{gc})^2}{9^2}}=0
X+40-20 \log 3-20 \log \sqrt{1+3^2}-20 \log \sqrt{1+\left ( \frac{3}{9} \right )^2}=0
X=-20dB
So, the compensator provides an attenuation od 20dB. Hence option (D) is correct.
T(s)=\frac{900}{s(s+1)(s+9)}
Phase crossover frequency of uncompensated system =(\omega _{pc}), at this frequency phase of T(j \omega ) is -180^{\circ}
Put s=j\omega \text{ in } T(s)
\begin{aligned} T(j\omega )&=\frac{900}{j\omega (j\omega +1)(j\omega +9)} \\ \angle T(j\omega )&=-90^{\circ}-tan ^{-1}\omega -tan^{-1}\left ( \frac{\omega }{9} \right ) \\ \text{At, } & (\omega _{pc})_1,\angle T(j\omega )=-180^{\circ}\\ -180^{\circ}&=-90^{\circ}-tan^{-1}(\omega _{pc})_1 -tan^{-1}\left ( \frac{(\omega _{pc})_1}{9} \right )\\ -90^{\circ}&=tan^{-1}\frac{(\omega _{pc})_1+\frac{(\omega _{pc})_1}{9}}{1-\frac{(\omega _{pc})^2_1}{9}} \\ \Rightarrow 1-\frac{(\omega _{pc})^2_1}{9}&=0 \\ \Rightarrow (\omega _{pc})_1&= 3 \text{ rad/sec} \end{aligned}
Gain cross frequency of compensated system, (\omega _{gc})_2= phase cross frequency of uncompensated system, (\omega _{pc})_1
\Rightarrow \;\;(\omega _{gc})_2=(\omega _{gc})_1=\; 3 rad/sec
Phase-margin =180^{\circ}+\angle T(j\omega )|_{\omega =(\omega _{gc})_2}
\Rightarrow \; \; \; 45^{\circ}=180^{\circ}+\angle T(j\omega )|_{\omega =(\omega _{pc})_2}
At (\omega _{gc})_2=3 rad/sec,
phase angle of \angle T(j\omega ) is -135^{\circ} and phase of uncompensated systen is -180^{\circ} at 3 rad/sec.
Therefore, the compensator provides phase lead of 45^{\circ} at the frequency of 3 rad/sec.
Let X dB is the gain provided by the compensator, so at gain cross frequency, |T(j\omega )|_{com}=1 or 0 dB. Gain of uncompensated system =|T(j\omega )|_{un-com} =\frac{100}{\omega \sqrt{1+\omega ^2}\sqrt{1+\left ( \frac{\omega }{9} \right )^2}}
|T(j\omega )|_{un-com} in dB
\;=40-20 \log \omega -20 \log \sqrt{1+\omega ^2} -20 \log \sqrt{1+\left ( \frac{\omega }{9} \right )^2}
Gain of compensated system,
|T(j\omega )|_{com}=X+|T(j\omega )_{un-com}|
|T(j\omega )|_{com} must be zero at gain cross frequency (\omega _{gc})_2
|T(j\omega_{gc} )_2|_{com}=X+40-20 \log (\omega _{gc})_2 -20 \log \sqrt{1+(\omega _{gc})^2_2}-20 \log \sqrt{1+\frac{(\omega _{gc})^2}{9^2}}=0
X+40-20 \log 3-20 \log \sqrt{1+3^2}-20 \log \sqrt{1+\left ( \frac{3}{9} \right )^2}=0
X=-20dB
So, the compensator provides an attenuation od 20dB. Hence option (D) is correct.
Question 8 |
A lead compensator used for a closed loop controller has the following transfer
function \frac{K(1+\frac{s}{a})}{1+\frac{s}{b}}
For such a lead compensator
For such a lead compensator
a \lt b | |
a \gt b | |
a \gt Kb | |
a \lt Kb |
Question 8 Explanation:
Transfer function=\frac{k\left ( 1+\frac{s}{a} \right )}{\left ( 1+\frac{s}{b} \right )}
Zero of TF=-a
Pole of TF=-b
For a lead-compensator, the zero is nearer to origin as compared to pole, hence the effect of zero is dominant, therefore, the lead-compensator when introduced in series with forward path of the trnasfer function the phase shift is increased.
So, from pole-zero configuration of the compensator a \lt b.
Zero of TF=-a
Pole of TF=-b
For a lead-compensator, the zero is nearer to origin as compared to pole, hence the effect of zero is dominant, therefore, the lead-compensator when introduced in series with forward path of the trnasfer function the phase shift is increased.
So, from pole-zero configuration of the compensator a \lt b.
There are 8 questions to complete.